brute compressor

Whim, with two threads that seem to optimize some violenceFor example, the last point of the plane to this topic, the point copy into 2 copiesOne is sorted by X, one is sorted by Y.Then double-threading violence, one processing x, one processing yIf the data uses X to decrement the card, then because of the double thread, it can't hold YIf the data uses Y to decrement the card, then the card can't hold XSo violent n^2 can be done.#include #include#include#include#includeusing namespacestd;struct

Test instructions: Given n points, your task is to make them all connected. You can create some new side, the cost equals two points distance squared (of course the smaller the better), there are several "package", you can buy, you buy, then some side can connect up,Each "package" is also spent, which allows you to find the minimum cost.Analysis: The first thought is to consider all the circumstances of the calculation, and then find the least, first Count no "package", and then some, with binar

The main idea: a ring can open and close. Now there are N (1Title Analysis: The set of rings to be opened with a binary number, a total of 2^n cases, enumeration of each case. When the ring is opened, the ring is isolated, the next is to judge the rest of the ring is connected with a few rings, if some ring is still connected with more than two rings, the scheme is not feasible, it is impossible to form a chain, and then determine whether the remaining ring is connected to a circle, if there is,

Because we're going to be poor here. The password includes 0-9,a-z,a-z a total of 62 characters, so we use 62 binary to traverse.First, we implement a 10 binary to 62 binary method. private static char[] CharSet = "0123456789abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ". ToCharArray (); private static string[] CharSet = {"0", "1", "2", "3", "4", "5", "6", "7", "8", "9",// "A", "B", "C", "D", "E", "F",

ABS written fabs led WA to a pitch.About Next_permutation () the basic application of the function in STLHttp://www.cnblogs.com/luosuo10/p/5479188.html1#include 2#include 3#include 4#include 5#include 6 using namespacestd;7 intt,n,a,b;8 inta[ -];9 Chars[ -];Ten Const intinf=0x3f3f3f; One voidinput () A { -n=0; - gets (s); the intlen=strlen (s); - - for(intI=0; i) - { + if('0''9') - { +a[n++] = s[i]-'0'; A } at } - - } - voidSolve () - { - if(n==2) i

Transferred from: http://www.zhengdazhi.com/archives/5631, using iptables now ssh per minute number of connectionsAllow local loopback interface accessIptables-a input-i lo-j ACCEPTRelease all links that have been establishedIptables-a input-m state–state established-j ACCEPTOnly two new connections per minute to SSH are allowed, no limit on established connections A 2 2 -M state–state new-j ACCEPTAdd default policy Deny allIptables-p INPUT DROP2. Use Denyhost to deny access to the IP of the wro

3 Lessons to a This problem, and finally timed out, is also pretty spelling.I didn't do it. The main aspect is not to enumerate a subset of binary numbersHere's the code:for (int S0 = S; S0; S0 = (s0-1) S) { }Here S0 is a subset of s!The idea of the topic is to enumerate all the circumstances, attention to memory "said the problem learned a lot"#include 1354Mobile Computing (brute force, binary enumeration, simply heartless)

not.Sample Input3niconiconi~pettan,pettan,tsurupettanwafuwafuSample OutputYesyesnoAuthor: JIANG, KaiSource: The acm-icpc Asia Mudanjiang Regional first Round Title Link: http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemId=5350Question: Ask whether the remaining string after removing non-alphabetic characters satisfies the structure of Ababa or Ababcab, where ABC is a different substringTitle Analysis: First get the target string, and then enumerate the substrings A and B, which is used i

Test instructions: Given 13 tiles, ask if you are "listening" to the card, if it is output "listen" which one.Analysis: This problem, very obvious violence, is on the original foundation put on a card, see is not can Hu, idea is very simple, also better realize, the result is tle, has been tle, this unscientific ah ...Very hard to write out, unexpectedly tle ... Heartache. is the first to determine a card, and then the flush analysis, in fact, is to prune, that is, if there are 1 or two cards, a

) Continue; intNextdoor =i; if((Perlights[nextdoor] curlights) = =0) //the lights are off and cannot be entered. Continue; if(States[nextdoor][curlights] = =1) //I've seen it, can't go in Continue; //the light is on, it can go inNode NewNode; Newnode.cur=Nextdoor; Newnode.curstepnum= Curstepnum +1; Newnode.lights=curlights; Newnode.step=Node.step; stringdesc ="-Move to room"+ std::to_string (Nextdoor +1); Desc+=".";

1. Environment CentOS 72, online installationYum Install-y epel-releaseYum Install-y Fail2banFail2ban structure/etc/fail2ban # # Fail2ban Service configuration directory/ETC/FAIL2BAN/ACTION.D # # iptables, mail and other action files directory/ETC/FAIL2BAN/FILTER.D # # condition matching file directory, filtering log key content/ETC/FAIL2BAN/JAIL.CONF # # Fail2ban Protection configuration file/etc/fail2ban/fail2ban.conf # # Fail2ban profile, defining log levels, logs, sock file locations, and mo

crackedSmbService login account number and passwordusing the command:Hydra 127.0.0.1-l root-p p.txt SMB4 crackedPop3Service login account number and passworduse command:hydra-l root-p p.txt my.pop3.mail POP35 crackedImapService login account number and passworduse command:hydra-l root-p p.txt 192.168.1.30 IMAP6 crackedHttp-proxyService login account number and passworduse command:hydra-l root-p p.txt http-proxy://192.168.1.207 crackedRdpService login account number and passworduse command

[1559] Jump to the Top of Mountain Time limit: Ms Memory limit: 65535 K Problem description Have you played a game named Minecraft?In the game, there are a thing alloy you can not destroy, and their shape is a 1*1*1 area cube. I can only jump 1 meter at vertical height, it means I can jump from x meters height to y meters height when y isn't more than X+1.Now, there are a rectangle field of m*n areas, each area is a positive interger x, x means the number of alloy . You can have only a jum

Test instructionsGive you a sequence of length n, M inquiry, one number per learningLet you answer how many pairs of products in a sequence are not less than it.Ideas:Preprocessing the number of pairs that are not more than the current value in the current sequence, and then subtracting the number from the previous one is the answer./************************************************author:d evil*********************************************** * */#include#include#include#include#include#include#in

not even up, so also to meet this condition, open the number of rings greater than or equal to need to join the fragment minus 1. The rest is how to judgeLooping and branches greater than two, first say how to determine how to calculate the branch is greater than two, in the event of violence, can be calculated separately except to open the ring and the number of rings connected to it, if more than 2,Then is the branch is greater than 2, and how to judge the ring, it is clear that with Dfs, fro

Title Link: Http://www.codeforces.com/problemset/problem/268/ATest instructions: If the home team's home shirt is the same as the away shirt in the football match, then ask the team to put on their jerseys, now there are N teams, each team's home shirt and the color of the go shirt tell you, they are 22 more than a game. The number of matches where the home team wears away jerseys.C + + code:#include using namespacestd;Const intMAXN = -;intX[MAXN], Y[MAXN], n, ans;intMain () {CIN>>N; for(inti =

Title Link: Http://www.codeforces.com/problemset/problem/327/ATest instructions: You now have n cards, these are 0 on one side and 1 on the other side. Numbering from 1 to n, you need to flip the [i,j] interval of the card once, so that the number of cards seen is 1 of the largest.C + + code:#include using namespacestd;Const intMAXN = the;intN, A[maxn], SUM[MAXN];intFlipintLintR) { intA1 = Sum[r]-sum[l-1]; intA2 = R-l +1-A1; //cout returnSum[n]-sum[r] + sum[l-1] +A2;}intMain () {CIN>>N;

; Alchild[m+1]=i;rchild[m+1]=J; +wight[m+1]=wight[i]+Wight[j]; theSearch (cnt-1, m+1); -vis[m+1]=-1; $ } thevis[j]=0; the } the } thevis[i]=0; - } in } the } the intMain () { About //freopen ("Input.txt", "RB", stdin); the //freopen ("Output.txt", "WB", stdout); the intN; thescanf"%d",N); + while(n--){ - init (); thescanf"%lf%d",r,s);Bayi the for(intI=1; i){ thescanf"%d",wight[i]); -vis[i]=0; - } the

Two days ago, after the myBB Forum exploit disappeared, the password cracked was abnormal ~ It is in the form of md5 (salt). md (pass ~ No progress ~ Today I wrote a small exploitation program ~ Hope everyone can use it ~ Pass.txt is the password dictionary ~~ One line ~ In addition, although the program uses a very junk Syntax Two days ago, after the myBB Forum exploit disappeared, the password cracked was abnormal ~ It is in the form of md5 (salt). md (pass ~No progress ~ Today I wrote a small

} - } -flag=1; - returnFlag; + } - + A intFunChar* A1,Char* B1,intl) at { - intk=0; - if(Compare (a1,b1,l)) -k=1; - if(l%2==1) - { in if(k==1) - return 1; to Else + return 0; - } the Else * { $ if(k==1)Panax Notoginseng return 1; - Else the { +l=l/2; A if((a1,b1,l) Fun (a1+l,b1+l,l)) | | (a1,b1+l,l) Fun (a1+l,b1,l)) ) the return 1; + E